Given a smooth curve $\alpha :I\to \mathbb{R}^n$,define arc length map $s:I\to \mathbb{R}$ as: $$s(t) = \int_{t_0}^t |\alpha'(u)|du$$
Show that if $|a'(u)| \ne 0$ ,map s on $I \to s(I)$ is a diffeomorphism.
I can prove that $s \in C^{\infty}$ by calculation.and injective since if$s(t_1) = s(t_2)$ then $\int_{t_1}^{t_2}|\alpha'|du = 0$ with $|a'(u)|>0$ so $t_1 = t_2$.
What I don't know is how to consider the inverse of $s$
My idea is using inverse function theorem,so inverse function is $C^1$ with inverse derivative being explicitly expressed,so $C^n$ is given by calculation?
What I believe you're missing is an induction argument. Let $g=s^{-1}$. By the Inverse Function Theorem, $$g'(u) = \frac 1{s'(g(u))}.$$ Since $s\in C^1$, we see that $g'$ is continuous and so $g\in C^1$. Now suppose you've shown that $s\in C^k$ implies that $g\in C^k$. Use this formula to deduce that if $s\in C^{k+1}$, then $g\in C^{k+1}$.